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Products of complex projective spaces are linearly independent in rational oriented bordism

Statement

Assume AC (The Axiom of Choice), inherited from the characteristic-number and projective-space suppliers. Let $k \geq 1$ and let $P_{J} = {C P}^{2 j_{1}} \times \dots \times {C P}^{2 j_{r}}$ range over the products of complex projective spaces indexed by the partitions $J$ of $k$ . Then the classes $[P_{J}]$ are linearly independent in the rational oriented bordism group $Ω_{4 k}^{S O} \otimes Q$ ; equivalently, if $\sum_{J} c_{J} [P_{J}] = 0$ with $c_{J} \in Q$ then all $c_{J} = 0$ . Consequently $\dim_{Q} (Ω_{4 k}^{S O} \otimes Q) \geq p (k),$ the number of partitions of $k$ .

Facts & Assumptions

Given: An integer $k \geq 1$ , the partitions $J$ of $k$ , the classes $[P_{J}] \in Ω_{4 k}^{S O}$ of the projective-space products with their product complex orientations, and rational coefficients $c_{J}$ .

[F1]

Characteristic numbers are cobordism invariants: oriented-cobordant closed oriented $4 k$ -manifolds have equal Pontryagin numbers, so for each partition $I$ of $k$ the Pontryagin number is a well-defined function on the oriented bordism classes of Unoriented and oriented bordism groups.

[F2]

Pontryagin numbers of a closed oriented manifold defines the number componentwise over the connected components, so $p_{I} [M ⊔ N] = p_{I} [M] + p_{I} [N]$ ; the group operation on $Ω_{4 k}^{S O}$ is $[M] + [N] = [M ⊔ N]$ and the classes of products are the well-defined products in Cartesian product makes bordism a graded ring (Unoriented and oriented bordism groups). Hence $p_{I}$ is additive for the group operation, $p_{I} ([M] + [N]) = p_{I} [M] + p_{I} [N]$ .

[F3]

Products of complex projective spaces have an invertible Pontryagin-number matrix: the ordinary Pontryagin-number matrix $A_{I, J} = ⟨ p_{I} (T P_{J}), [P_{J}] ⟩$ , with rows and columns indexed by the partitions of $k$ , is invertible over $Q$ ; Characteristic numbers of products satisfy the Whitney-sum and Kunneth product formulas identifies the numbers of the products with the evaluations used in that matrix.

Proof

1.1F1F2

Each Pontryagin number is additive. By [F1], $p_{I}$ is constant on oriented cobordism classes, hence well defined on $Ω_{4 k}^{S O}$ ; by the componentwise definition and the disjoint-union group law [F2], $p_{I} [M ⊔ N] = p_{I} [M] + p_{I} [N]$ for closed oriented $4 k$ -manifolds, so $p_{I} : Ω_{4 k}^{S O} \to Z$ is a group homomorphism. Extending scalars, $(z \otimes q) \mapsto q p_{I} [z]$ is $Z$ -balanced and $Q$ -bilinear, hence induces a linear functional $p_{I} \otimes Q : Ω_{4 k}^{S O} \otimes Q \to Q$ ; the balancing relation $p_{I} [n z \otimes q] = n q p_{I} [z] = p_{I} [z \otimes n q]$ holds by the definition of the tensor product.

2.1F2F3step 1.1

Suppose $\sum_{J} c_{J} [P_{J}] = 0$ in $Ω_{4 k}^{S O} \otimes Q$ . Applying the functional $p_{I} \otimes Q$ for every partition $I$ of $k$ and using additivity [F2] gives $\sum_{J} A_{I, J} c_{J} = 0$ for every $I$ , that is, the matrix equation $A c = 0$ with $A_{I, J} = ⟨ p_{I} (T P_{J}), [P_{J}] ⟩$ as in [F3].

3.1F3step 2.1

By [F3] the matrix $A$ is invertible over $Q$ , so $A c = 0$ forces $c = 0$ : all coefficients vanish and the classes $[P_{J}]$ are linearly independent in $Ω_{4 k}^{S O} \otimes Q$ . A linearly independent family of $p (k)$ vectors in a vector space gives the lower bound $\dim_{Q} (Ω_{4 k}^{S O} \otimes Q) \geq p (k)$ .

4.1F3step 3.1∎

Boundary and convention remarks. For $k = 0$ , Zero-dimensional bordism groups gives the positive point generator; there is one partition, the empty one, with $P_{\emptyset}$ a point, $Ω_{0}^{S O} \otimes Q ≅ Q$ on the positive point class, and $p (k) = 1$ ; the lower bound is still correct and no independence beyond a nonzero class is claimed, but the statement is formulated for $k \geq 1$ where the products have positive dimension. The argument gives only a lower bound: it neither produces a rational Hurewicz theorem nor any spanning family, and no upper bound on $Ω_{4 k}^{S O} \otimes Q$ is asserted. The number $p (k)$ counts the partitions of $k$ , and the index set of [F3] is exactly that finite set. No choice beyond the cited suppliers is used.

Depends on

Used by

Products of complex projective spaces span rational oriented bordism Proposition

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Sources

John Milnor and James Stasheff, Characteristic Classes (original pagination; chapters 16-18 of the re-typeset scan) (standard reference, not scraped)
Daniel S. Freed, Bordism: Old and New (lecture notes, UT Austin, Fall 2012) (standard reference, not scraped)